Question

Liquid bromine \(\left(\mathrm{Br}_{2}\right)\) is spilled in a laboratory accident and evaporates. Assuming the vapor behaves as an ideal gas, with a temperature \(20.0^{\circ} \mathrm{C}\) and a pressure of \(101.0 \mathrm{kPa}\), find its density.

Short answer

Answer: The density of bromine gas at \(20.0^{\circ} \mathrm{C}\) and \(101.0 \mathrm{kPa}\) is approximately \(5.52 \, g/L\).

Step-by-step solution

Convert the temperature to Kelvin

First, we need to convert the given temperature from Celsius to Kelvin, as the Ideal Gas Law works with temperatures in Kelvin. To do this, we add 273.15 to the Celsius temperature: \(T(K) = T(^\circ C) + 273.15\) \(T(K) = 20.0 + 273.15\) \(T(K) = 293.15 K\)

Calculate the molar mass of bromine

To find the mass of the gas, we need to know the molar mass of bromine. Bromine has a molar mass of approximately 79.9 g/mol for each bromine atom, and since it is diatomic (Br₂), we have: \(Molar mass (Br_2) = 2 \times 79.9 = 159.8 \, g/mol\)

Rearrange the Ideal Gas Law to find moles of bromine

We will rearrange the Ideal Gas Law formula to find the moles of bromine: \(n = \frac{PV}{RT}\) And we'll plug in the given values: P = 101.0 kPa = 101000 Pa (we convert from kPa to Pa) R = 8.314 J/mol K (gas constant) T = 293.15 K (from Step 1) \(n = \frac{(101000 \, Pa)(V)}{(8.314 \, J/mol \cdot K)(293.15 \, K)}\) Note that we now have the equation in terms of V (volume).

Express mass as a function of the number of moles and molar mass

Using the molar mass of bromine, we can find the mass of the gas: \(mass = n \times Molar mass\) \(mass = n \times 159.8 \, g/mol\)

Find the density of bromine gas

We can now express the density as a function of volume, mass, moles, and molar mass: \(density = \frac{mass}{volume}\) Plugging in our expressions for mass and n from the previous steps: \(density = \frac{n \times 159.8 \, g/mol}{V}\) But we know that n = \(\frac{PV}{RT}\), so we'll replace that: \(density = \frac{\frac{PV}{RT} \times 159.8 \, g/mol}{V}\) Now we can cancel out V: \(density = \frac{P \times 159.8 \, g/mol}{RT}\) Finally, plug in the values for P, R, and T: \(density = \frac{(101000 \, Pa) \times 159.8 \, g/mol}{(8.314 \, J/mol \cdot K)(293.15 \, K)}\) \(density \approx 5.52 \, g/L\) So the density of bromine gas at \(20.0^{\circ} \mathrm{C}\) and \(101.0 \mathrm{kPa}\) is approximately \(5.52 \, g/L\).

Key concepts

Molar Mass

Molar mass is a fundamental characteristic of a chemical element or compound and signifies the mass of one mole of that substance. It's expressed in grams per mole (g/mol) and is calculated by summing the atomic masses of all atoms in a molecule. For instance, liquid bromine (\text{Br}_2), as in our exercise, is composed of two bromine atoms. Since the atomic mass of a single bromine atom is approximately 79.9 amu (atomic mass units), the molar mass of \text{Br}_2 would be the mass of two bromine atoms, or:
\[ Molar \text{ mass } (Br_2) = 2 \times 79.9 = 159.8 \text{ g/mol} \]
The molar mass is crucial in determining the density of a gas since it relates the mass of the gas to the number of moles, a key step in the calculation process.

Temperature Conversion

To work with the Ideal Gas Law, temperature must be expressed in Kelvin (K), the SI base unit for thermodynamic temperature. Temperature conversions are essential because the various scales relate to different reference points. Celsius, for example, defines the freezing and boiling points of water as 0 and 100 degrees, respectively, while Kelvin starts at absolute zero, where particles theoretically cease motion.
To convert Celsius to Kelvin, one adds 273.15 to the Celsius temperature:
\[ T(K) = T(^\text{\circ}C) + 273.15 \]
In the case of bromine gas, the lab temperature is given as:
\[ T(K) = 20.0 + 273.15 = 293.15 \text{ K} \]
This precise conversion is necessary for calculations involving gas behavior and density.

Gas Constant

The gas constant, symbolized by R, is a key component in the Ideal Gas Law equation and is the same for all ideal gases. Its value is approximately:\[R = 8.314 \text{ J/mol} \text{ K} \]. It represents the proportionality constant that relates the energy scale to the thermodynamic temperature scale, when dealing with an amount of substance in moles.

To illustrate its application, in the density calculation for bromine gas, the gas constant R is used to relate pressure (P), volume (V), temperature (T), and the number of moles of gas (n) in the equation:\[n = \frac{PV}{RT}\]. It is crucial to understand that this constant is vital in transforming the theoretical relationships into practical calculations that can produce useful results, such as determining the density of a gas under specific conditions.

Moles of Gas

Moles of gas, denoted as n, is a measure of the quantity of a substance. It's a count of how many molecules, atoms, or formula units are present. One mole is equivalent to Avogadro's number (\text{6.022} \times \text{10}^{23}) of particles.

In gas calculations, understanding moles is indispensable. For example, the Ideal Gas Law utilizes moles to link physical characteristics of the gas, such as pressure and volume, to the amount of the substance. By rearranging the Ideal Gas Law, one can solve for the moles of the gas when the pressure, volume, and temperature are known:\[n = \frac{PV}{RT}\].

This relationship allows us to calculate important properties of gases, including density, by knowing how many moles occupy a given volume. It's important to comprehend that while volume and pressure can vary largely with changes in conditions, the mole quantity provides a consistent basis for comparison and calculation of gas properties.